The functional equation for Riemann's zeta
May 17, 2015. A self-contained proof of the functional equation for Riemann’s zeta function, mainly written for my own edification.
Introduction
Prerequisites: first course in complex analysis, the gamma and beta function.
You often hear in undergraduate courses that a complex function can be extended by analytic continuation to a larger domain. It sounds like a messy, technical business, and the details are usually “beyond the scope” of the context of utterance. What you don’t hear is that analytic continuation usually works by finding a beautiful formula relating values in the old domain to the new, larger domain, or which works in the old domain but is valid in the new. Loosely speaking, this uniquely determines the function on the larger domain because analytic functions are “rigid”.
My favourite example is the functional equation for the Riemann zeta function $\zeta(s)$, proved by Riemann in his remarkable 1859 paper:
or if we define $\Xi(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s)$, then $\Xi(s) = \Xi(1-s)$. This elegant, symmetric identity is not too hard to derive, and we will do so below. If we know the values of $\zeta(s)$ for $\Re(s) > 0$, we can use this equation to find the values everywhere (except the singularity at $s = 1$). As it turns out, the functional equation tells us much more!
A small extension and the gamma function
Let's label the parts of our complex argument $s := \sigma + i\tau$. For $\sigma > 1$, we can define $\zeta(s)$ by the series
\[
\zeta(s) = \sum_{n\geq 1} \frac{1}{n^s}.
\]This means that
\[
\zeta(s) = \sum_{n\geq 1} \frac{1}{n^s}.
\]This means that
\[
\begin{align}
1 - 2^{-s}+3^{-s}-4^{-s}+\cdots & = (1 + 2^s + 3^2 + \cdots) - 2(2^s + 4^s + \cdots)\\
& = \zeta(s)-2\sum_{n\geq 0}(2n)^{-s} \\ & = (1-2^{1-s})\zeta(s).
\end{align}
\]In other words,
\[
\zeta(s) = (1 - 2^{1-s})^{-1}\sum_{n\geq 1}(-1)^nn^{-s}.\label{ext}\tag{1}
\]The series on the right is alternating and converges for $\sigma > 0$. So, this expression analytically continues $\zeta(s)$ to $\sigma > 0$ except at $s = 1$, where the prefactor $(1 - 2^{1-s})^{-1}$ blows up.
\begin{align}
1 - 2^{-s}+3^{-s}-4^{-s}+\cdots & = (1 + 2^s + 3^2 + \cdots) - 2(2^s + 4^s + \cdots)\\
& = \zeta(s)-2\sum_{n\geq 0}(2n)^{-s} \\ & = (1-2^{1-s})\zeta(s).
\end{align}
\]In other words,
\[
\zeta(s) = (1 - 2^{1-s})^{-1}\sum_{n\geq 1}(-1)^nn^{-s}.\label{ext}\tag{1}
\]The series on the right is alternating and converges for $\sigma > 0$. So, this expression analytically continues $\zeta(s)$ to $\sigma > 0$ except at $s = 1$, where the prefactor $(1 - 2^{1-s})^{-1}$ blows up.
The gamma function is the workhorse of the special functions. We will need two of its properties: the reflection formula
\[
\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin \pi s}.\tag{2}\label{gamma}
\]and the duplication formula
\[
\Gamma(s)\Gamma(s + \tfrac{1}{2}) = 2^{1-2s}\sqrt{\pi}\Gamma(2s).\tag{3}\label{gamma2}
\]These are proved in appendix B.
\[
\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin \pi s}.\tag{2}\label{gamma}
\]and the duplication formula
\[
\Gamma(s)\Gamma(s + \tfrac{1}{2}) = 2^{1-2s}\sqrt{\pi}\Gamma(2s).\tag{3}\label{gamma2}
\]These are proved in appendix B.
Proving the functional equation
Let's proceed with the proof. First, we use induction to prove the identity
\begin{align}
(1-e^{-x2^{-n}})\prod_{k=1}^{n}(1+e^{-x2^{-k}}) = 1-e^{-x}\label{exp}\tag{4}
\end{align}for $x > 0$. Clearly, the formula holds for $n = 0$. Assume it holds for $n$. Then
\begin{align}
(1-e^{-x2^{-(n+1)}})\prod_{k=1}^{n+1}(1+e^{-x2^{-k}}) & = (1-e^{-x2^{-(n+1)}})(1+e^{-x2^{-(n+1)}})\prod_{k=1}^{n}(1+e^{-x2^{-k}}) \\
& = (1 - e^{-x2^{-n}}) \cdot\frac{(1-e^{-x})}{(1 - e^{-x2^{-n}}) } = 1-e^{-x}.
\end{align}Take logarithms and differentiate both sides of $(\ref{exp})$ with respect to $x$:
\begin{align}
\frac{2^{-n}e^{-x2^{-n}}}{1-e^{-x2^{-n}}} + \sum_{k=1}^{n}\frac{-2^{-k}e^{-x2^{-k}}}{1+e^{-x2^{-k}}} = \frac{e^{-x}}{1 - e^{-x}}.
\end{align}We can rearrange this to get
\begin{align}
\sum_{k=1}^{n}\frac{2^{-k}}{e^{x2^{-k}}+1} = \frac{2^{-n}}{e^{x2^{-n}}-1} -\frac{1}{e^x -1} = \frac{1}{x}\frac{x2^{-n}}{e^{x2^{-n}}-1} -\frac{1}{e^x -1}.
\end{align}For fixed $x$, as $n\to\infty$, $x2^{-n} \to 0$ and we can Taylor expand the exponential:
\begin{align}
\frac{x2^{-n}}{e^{x2^{-n}}-1} \approx \frac{x2^{-n}}{1 + x2^{-n} - 1} = 1.
\end{align}Hence, as $n\to\infty$,
\begin{align}
\sum_{k\geq 1}\frac{2^{-k}}{e^{x2^{-k}}+1} = \frac{1}{x} -\frac{1}{e^x -1}. \tag{5}\label{sum}
\end{align}Consider $0 < \sigma < 1$. We now multiply both sides of $(\ref{sum})$ by $-x^{s-1}$, integrate, and simplify the result with $(\ref{ext})$:
\begin{align}
\int_{0}^\infty \left(\frac{1}{e^x -1}-\frac{1}{x}\right)x^{s-1}\,dx & = \int_0^\infty \sum_{k\geq1}\frac{-2^{-k}}{e^{x2^{-k}}+1} \,x^{s-1}\,dx\\
&= \int_0^\infty \sum_{k\geq1}\frac{-e^{-x2^{-k}}2^{-k}}{1+e^{-x2^{-k}}} \,x^{s-1}\,dx\\
& = \sum_{k\geq 1}2^{-k}\sum_{n\geq 1}(-1)^n\int_0^\infty e^{-nx2^{-k}}x^{s-1}\,dx\\
& = \sum_{k\geq 1}2^{-k}\sum_{n\geq 1}(-1)^n\int_0^\infty e^{-t}t^{s-1}(2^kn^{-1})^{s}\,dt\\
& = \Gamma(s)\sum_{k\geq 1}2^{(s-1)k}\sum_{n\geq 1}(-1)^nn^{-s}\\
& = \Gamma(s)\frac{2^{s-1}}{1-2^{s-1}}(2^{1-s}-1)\zeta(s) \\ & = \Gamma(s)\zeta(s).
\end{align}Note that we were able to interchange sums and integral from absolute (hence uniform) convergence, and we expanded the left side of $(\ref{sum})$ as a geometric series. This completes the first part of the proof.
The second is less elementary. First of all, make the change of variables $x = \sqrt{2\pi} y$ in the first integral and use it to define a new function $f$:
\begin{align}
(2\pi)^{s/2}\int_{0}^\infty \left(\frac{1}{e^{\sqrt{2\pi}y} -1} - \frac{1}{\sqrt{2\pi}y}\right)y^{s-1}\,dy := (2\pi)^{s/2}\int_{0}^\infty f(y) y^{s-1}\,dy.\tag{6}\label{f(t)}
\end{align}But this is equal to $\Gamma(s)\zeta(s)$, so
\[
F(s) := (2\pi)^{-s/2}\Gamma(s)\zeta(s) = \int_{0}^\infty f(y) y^{s-1}\,dy.
\]At this point, we assert that $f(t)$ has the property of being equal to its own Fourier sine transform:
\[
f(t) = \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin(tx)\,dx. \tag{7}\label{sine}
\]For a proof, see appendix A. This nice property almost immediately gives us the functional equation. First, we need an integral (see appendix B):
\begin{align}
\int_0^\infty y^{s-1} \sin(xy)\,dy & = x^{-s}\Gamma(s)\sin(\pi s/2).
\end{align}We can use this to derive a reflection formula for $F$:
\begin{align}
F(s) = \int_{0}^\infty f(y) y^{s-1}\,dy & = \sqrt{\frac{2}{\pi}}\int_{0}^\infty \left(\int_0^\infty f(x)\sin(xy)\,dx\right)y^{s-1}\,dy \\
& = \sqrt{\frac{2}{\pi}}\int_{0}^\infty f(x) \left(\int_0^\infty y^{s-1} \sin(xy)\,dy\right)dx \\
& = \sqrt{\frac{2}{\pi}}\Gamma(s)\sin(\pi s/2)\int_{0}^\infty f(x)x^{-s}\, dx \\
& = \sqrt{\frac{2}{\pi}}\Gamma(s)\sin(\pi s/2)F(1-s).
\end{align}The last line lets us continue $F$ outside the strip $0 < \sigma < 1$. We can rewrite this reflection identity more explicitly as
\begin{align}
(2\pi)^{-s/2}\zeta(s)\Gamma(s)&=\sqrt{\frac{2}{\pi}}\Gamma(s)\sin(\pi s/2)\cdot (2\pi)^{(s-1)/2}\Gamma(1-s)\zeta(1-s)\\
& =2^{s/2}\pi^{-s/2}\frac{\sin(\pi s/2)}{\pi}\Gamma(s)\Gamma(1-s)\zeta(1-s).
\end{align}We can use the following consequences of the reflection and duplication formula to simplify further:
\[
\frac{\sin(\pi s /2)}{\pi} = \frac{1}{\Gamma(s/2)\Gamma(1 - s/2)}, \quad \frac{\Gamma(1-s)}{\Gamma(1-s/2)} = (2\pi)^{-s}\Gamma((1-s)/2).
\](See appendix B for more detail.) With the help of these identities, we finally obtain
\begin{align}
(2\pi)^{-s/2}\Gamma(s/2)\zeta(s) & = 2^{-s}\pi^{-1/2}\cdot (2\pi)^{s/2}\Gamma((1-s)/2)\zeta(1-s)\\
& = 2^{-s/2}\pi^{(s-1)/2}\Gamma((1-s)/2)\zeta(1-s).
\end{align}Multiplying out $2^{-s/2}$, we see this is equivalent to the functional equation:
\[
\pi^{-s/2}\Gamma(s/2)\zeta(s) = \pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).
\]It is a thing of beauty. But it's very useful too. Not only we can use it to analytically continue $\zeta$, it immediately yields some interesting values. For instance, substitute $s = 2$:
\begin{align*}
\pi^{-1}\Gamma(1)\zeta(2) & = \pi^{1/2}\Gamma(-1/2)\zeta(-1).
\end{align*}Since $\Gamma(-1/2) = -2\Gamma(1/2) = -2\sqrt{\pi}$, and $\zeta(2) = \pi^2/6$, we find that
\[
\zeta(-1) = \frac{\pi^{-3/2}\Gamma(1)\zeta(2)}{\Gamma(-1/2)} = \frac{\sqrt{\pi}}{-12\sqrt{\pi}} = -\frac{1}{12}.
\]
\[
\frac{1}{e^u - 1} =\frac{1}{u} - \frac{1}{2} + \sum_{k\geq 1}\frac{2u}{4k^2\pi^2 + u^2}.\tag{8}\label{sinlog}
\]To prove this, start with the product formula for sine,
\[
\sin z = z\prod_{k\geq 0}\left(1 -\frac{z^2}{\pi^2k^2}\right).
\]Set $z = u/2i$ and take log of both sides
\[
\log\left[e^{u/2} - e^{-u/2}\right] = \log(1 - e^{-u}) + \frac{u}{2} = \log(u) + \sum_{k\geq 0}\log\left(\frac{4\pi^2k^2 + u^2}{4\pi^2k^2}\right).
\]Differentiating with respect to $u$, we obtain $(\ref{sinlog})$. We can use this to simplify the integral of $(\ref{f(t)})$:
\begin{align*}
\int_0^\infty f(x) \sin(xt)\,dx & = \int_0^\infty \left(\frac{1}{e^{\sqrt{2\pi}x} -1} - \frac{1}{\sqrt{2\pi}x}\right) \sin(xt)\,dx\\
& =\sum_{k\geq 1}\int_0^\infty e^{-kx\sqrt{2\pi}}\sin(xt)\,dx -\frac{1}
{\sqrt{2\pi}}\int_0^\infty \frac{\sin(xt)}{x}\,dx \\
& =\sum_{k\geq 1}\frac{1}{t}\int_0^\infty e^{-(\sqrt{2\pi}k/t)\eta}\sin(\eta)\,d\eta - \frac{1}{4}\sqrt{2\pi} \\
& = \sum_{k\geq 1} \frac{t}{2\pi k^2+t^2}- \frac{1}{4}\sqrt{2\pi}\\
& = \frac{1}{2}\sqrt{2\pi}\left[\frac{1}{e^{\sqrt{2\pi}t} - 1} - \frac{1}{\sqrt{2\pi}t} + \frac{1}{2}\right] -\frac{1}{4}\sqrt{2\pi}\\
& = \sqrt{\frac{\pi}{2}}f(t),
\end{align*}which is equivalent to $(\ref{sine})$. We have used the integrals
\begin{align*}
\int_{0}^\infty \frac{\sin(xt)}{x}\,dx & = \frac{\pi}{2} \\
\int_{0}^\infty e^{-\lambda x}\sin(x)\,dx & = \frac{1}{\lambda^2 + 1}, \quad \lambda > 0.
\end{align*}
To prove equation $(\ref{gamma})$, namely
\[
\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin \pi s}
\]we recall a convenient way of writing the relation between the gamma functions and beta integral,
\[
\int_0^\infty \frac{t^{\alpha-1}}{(1+t)^{\alpha + \beta}}dt = B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}.
\]For $0 < \alpha < 1$, we set $\beta = 1 - \alpha$ to obtain
\[
\int_0^\infty \frac{t^{\alpha-1}}{1+t}dt = \frac{\Gamma(\alpha)\Gamma(1 - \alpha)}{\Gamma(1)} = \Gamma(\alpha)\Gamma(1 - \alpha).
\]It is now just a matter of doing the integral. Consider the contour integral
\[
\int_{C(R, \epsilon)} \frac{z^{\alpha - 1}}{1 - z} dz,
\]where $C(R, \epsilon)$ is the following contour:
The only singularity within the contour lies at $z = 1$. Provided we take the principal value of $z^{\alpha - 1}$, the residue is simply $-1^{\alpha - 1} = -1$. By the residue theorem,
\[
\int_{C(R, \epsilon)} \frac{z^{\alpha - 1}}{1 - z} dz = -2\pi i.
\]We split the integral up as in the picture, and note that along $C_1$, $z = Re^{i\theta}$; along $C_2$, $z = e^{i\pi}t$; along $C_3$, $z = z = \epsilon e^{i\theta}$; and along $C_4$, $z = e^{-i\pi}t$. (The $e^{\pm i\pi}$ is significant, and dictates how we take powers consistent with the principal value used to compute the residue.) Hence,
\begin{align}
-2\pi i & = \int_{C_1} \frac{z^{\alpha - 1}}{1 - z} dz + \int_{C_2} \frac{z^{\alpha - 1}}{1 - z} dz + \int_{C_3} \frac{z^{\alpha - 1}}{1 - z} dz + \int_{C_4} \frac{z^{\alpha - 1}}{1 - z} \\
& = \int_{-\pi}^\pi \frac{iR^{\alpha}e^{i\alpha\theta}}{1 - Re^{i\theta}} d\theta + \int_R^\epsilon \frac{t^{\alpha - 1}e^{i\alpha\pi}}{1 + t} dt + \int_{\pi}^{-\pi} \frac{i\epsilon^{\alpha}e^{i\alpha\theta}}{1 - \epsilon e^{i\theta}} d\theta + \int_\epsilon^R \frac{t^{\alpha - 1}e^{-i\alpha\pi}}{1 + t} dt \\
& = (e^{-i\alpha\pi} - e^{i\alpha\pi})\int_\epsilon^R \frac{t^{\alpha - 1}}{1 + t} dt +\int_{-\pi}^\pi \frac{iR^{\alpha}e^{i\alpha\theta}}{1 - Re^{i\theta}} d\theta + \int_{\pi}^{-\pi} \frac{i\epsilon^{\alpha}e^{i\alpha\theta}}{1 - \epsilon e^{i\theta}} d\theta.
\end{align}If we send $\epsilon \to 0$ and $R \to \infty$, the $C_1$ and $C_3$ angular integrals vanish, since the $C_1$ integrand $\sim R^\alpha / R = R^{\alpha-1}$ and the $C_3$ integrand $\sim \epsilon^\alpha$. Thus
\begin{align*}
2\pi i & = (e^{i\alpha\pi} - e^{-i\alpha\pi})\int_\epsilon^R \frac{t^{\alpha - 1}}{1 + t} dt \\
& = 2 i \sin(\alpha \pi)\int_\epsilon^R \frac{t^{\alpha - 1}}{1 + t} dt = 2 i \sin(\alpha \pi)\Gamma(\alpha)\Gamma(1 - \alpha).
\end{align*}The reflection formula $(\ref{gamma})$ follows for all complex $\alpha$ by analytic continuation. (For the singular points $\alpha \in \mathbb{Z}_{\leq 0}$, the Laurent series coefficients also match.)
2. The duplication formula
We now prove the duplication formula $(\ref{gamma2})$. First, we make the substitution $t = (1+u)/2$ in a more traditional expression for the beta function:
\begin{align}
B(z, z) = \frac{\Gamma(z)\Gamma(z)}{\Gamma(2z)} & = \int_0^1 t^{z - 1}(1-t)^{z-1} \,dt \\
&= \frac{1}{2}\int_{-1}^1 \left(\frac{1+u}{2}\right)^{z-1}\left(\frac{1-u}{2}\right)^{z-1} \,du\\
& =\frac{1}{2^{2z-1}}\int_{-1}^1 (1-u^2)^{z-1} \,du = \frac{1}{2^{2z-2}}\int_{0}^1 (1-u^2)^{z-1} \,du.
\end{align}But if we make the substitution $t = u^2$ in $B(1/2, z)$, we get
\begin{align}
B(\tfrac{1}{2}, z) = \frac{\Gamma(\tfrac{1}{2})\Gamma(z)}{\Gamma(z + \tfrac{1}{2})} & = \int_0^1 t^{1/2 - 1}(1-t)^{z-1}\,dt \\
& = \int_0^1 u^{- 1}(1-u^2)^{z-1}2u\,du \\
& = 2\int_0^1 (1-u^2)^{z-1}\,du.
\end{align}Equating the integrals yields
\[
2^{2z-1}\frac{\Gamma(z)\Gamma(z)}{\Gamma(2z)} = \frac{\Gamma(\tfrac{1}{2})\Gamma(z)}{\Gamma(z + \tfrac{1}{2})}.
\]Using $\Gamma(1/2) = \sqrt{\pi}$ and rearranging, we finally obtain $(\ref{gamma2})$:
\[
\Gamma(z)\Gamma(z+\tfrac{1}{2}) = 2^{1-2z}\sqrt{\pi}\Gamma(2z).
\] An integral
Finally, we sketch a proof that for $x > 0$,
\begin{align}
\int_0^\infty y^{s-1} \sin(xy)\,dy & = x^{-s}\Gamma(s)\sin(\pi s/2).
\end{align} Suppose that $0 < s < 1$. Making the change of variable $t = xy$,
\begin{align}
\int_0^\infty y^{s-1} \sin(xy)\,dy & = x^{-s}\int_0^\infty t^{s-1} \sin(t)\,dt\\
& = x^{-s}\Im \int_0^\infty t^{s-1} e^{it}\,dt.
\end{align}This looks like the gamma integral, except for the factor of $-i$ in the exponential. If we integrate around the boundary $C_R$ of the first quadrant of the disc of size $R$, we find from the residue theorem that
\begin{align*}
\int_0^R z^{s-1} e^{iz}\,dz + I(R) + \int_{iR}^0 z^{s-1} e^{iz}\,dz = 0,
\end{align*}where
\[
I(R) \sim R^{s-1}\int e^{-R\sin\theta} d\theta \to 0
\]as $R\to\infty$. Hence,
\begin{align*}
\int_0^\infty z^{s-1} e^{iz}\,dz &= \lim_{R\to\infty}\int_0^{iR} z^{s-1} e^{iz}\,dz\\
&= i^s\lim_{R\to\infty}\int_{0}^{R} x^{s-1} e^{-x}\,dx \\
&= e^{i\pi s/2}\Gamma(s).
\end{align*}Taking the imaginary component gives the desired result. By analytic continuation, the result is true for $s = \sigma + i\tau$, $0 < \sigma < 1$.
References
\begin{align}
(1-e^{-x2^{-n}})\prod_{k=1}^{n}(1+e^{-x2^{-k}}) = 1-e^{-x}\label{exp}\tag{4}
\end{align}for $x > 0$. Clearly, the formula holds for $n = 0$. Assume it holds for $n$. Then
\begin{align}
(1-e^{-x2^{-(n+1)}})\prod_{k=1}^{n+1}(1+e^{-x2^{-k}}) & = (1-e^{-x2^{-(n+1)}})(1+e^{-x2^{-(n+1)}})\prod_{k=1}^{n}(1+e^{-x2^{-k}}) \\
& = (1 - e^{-x2^{-n}}) \cdot\frac{(1-e^{-x})}{(1 - e^{-x2^{-n}}) } = 1-e^{-x}.
\end{align}Take logarithms and differentiate both sides of $(\ref{exp})$ with respect to $x$:
\begin{align}
\frac{2^{-n}e^{-x2^{-n}}}{1-e^{-x2^{-n}}} + \sum_{k=1}^{n}\frac{-2^{-k}e^{-x2^{-k}}}{1+e^{-x2^{-k}}} = \frac{e^{-x}}{1 - e^{-x}}.
\end{align}We can rearrange this to get
\begin{align}
\sum_{k=1}^{n}\frac{2^{-k}}{e^{x2^{-k}}+1} = \frac{2^{-n}}{e^{x2^{-n}}-1} -\frac{1}{e^x -1} = \frac{1}{x}\frac{x2^{-n}}{e^{x2^{-n}}-1} -\frac{1}{e^x -1}.
\end{align}For fixed $x$, as $n\to\infty$, $x2^{-n} \to 0$ and we can Taylor expand the exponential:
\begin{align}
\frac{x2^{-n}}{e^{x2^{-n}}-1} \approx \frac{x2^{-n}}{1 + x2^{-n} - 1} = 1.
\end{align}Hence, as $n\to\infty$,
\begin{align}
\sum_{k\geq 1}\frac{2^{-k}}{e^{x2^{-k}}+1} = \frac{1}{x} -\frac{1}{e^x -1}. \tag{5}\label{sum}
\end{align}Consider $0 < \sigma < 1$. We now multiply both sides of $(\ref{sum})$ by $-x^{s-1}$, integrate, and simplify the result with $(\ref{ext})$:
\begin{align}
\int_{0}^\infty \left(\frac{1}{e^x -1}-\frac{1}{x}\right)x^{s-1}\,dx & = \int_0^\infty \sum_{k\geq1}\frac{-2^{-k}}{e^{x2^{-k}}+1} \,x^{s-1}\,dx\\
&= \int_0^\infty \sum_{k\geq1}\frac{-e^{-x2^{-k}}2^{-k}}{1+e^{-x2^{-k}}} \,x^{s-1}\,dx\\
& = \sum_{k\geq 1}2^{-k}\sum_{n\geq 1}(-1)^n\int_0^\infty e^{-nx2^{-k}}x^{s-1}\,dx\\
& = \sum_{k\geq 1}2^{-k}\sum_{n\geq 1}(-1)^n\int_0^\infty e^{-t}t^{s-1}(2^kn^{-1})^{s}\,dt\\
& = \Gamma(s)\sum_{k\geq 1}2^{(s-1)k}\sum_{n\geq 1}(-1)^nn^{-s}\\
& = \Gamma(s)\frac{2^{s-1}}{1-2^{s-1}}(2^{1-s}-1)\zeta(s) \\ & = \Gamma(s)\zeta(s).
\end{align}Note that we were able to interchange sums and integral from absolute (hence uniform) convergence, and we expanded the left side of $(\ref{sum})$ as a geometric series. This completes the first part of the proof.
The second is less elementary. First of all, make the change of variables $x = \sqrt{2\pi} y$ in the first integral and use it to define a new function $f$:
\begin{align}
(2\pi)^{s/2}\int_{0}^\infty \left(\frac{1}{e^{\sqrt{2\pi}y} -1} - \frac{1}{\sqrt{2\pi}y}\right)y^{s-1}\,dy := (2\pi)^{s/2}\int_{0}^\infty f(y) y^{s-1}\,dy.\tag{6}\label{f(t)}
\end{align}But this is equal to $\Gamma(s)\zeta(s)$, so
\[
F(s) := (2\pi)^{-s/2}\Gamma(s)\zeta(s) = \int_{0}^\infty f(y) y^{s-1}\,dy.
\]At this point, we assert that $f(t)$ has the property of being equal to its own Fourier sine transform:
\[
f(t) = \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin(tx)\,dx. \tag{7}\label{sine}
\]For a proof, see appendix A. This nice property almost immediately gives us the functional equation. First, we need an integral (see appendix B):
\begin{align}
\int_0^\infty y^{s-1} \sin(xy)\,dy & = x^{-s}\Gamma(s)\sin(\pi s/2).
\end{align}We can use this to derive a reflection formula for $F$:
\begin{align}
F(s) = \int_{0}^\infty f(y) y^{s-1}\,dy & = \sqrt{\frac{2}{\pi}}\int_{0}^\infty \left(\int_0^\infty f(x)\sin(xy)\,dx\right)y^{s-1}\,dy \\
& = \sqrt{\frac{2}{\pi}}\int_{0}^\infty f(x) \left(\int_0^\infty y^{s-1} \sin(xy)\,dy\right)dx \\
& = \sqrt{\frac{2}{\pi}}\Gamma(s)\sin(\pi s/2)\int_{0}^\infty f(x)x^{-s}\, dx \\
& = \sqrt{\frac{2}{\pi}}\Gamma(s)\sin(\pi s/2)F(1-s).
\end{align}The last line lets us continue $F$ outside the strip $0 < \sigma < 1$. We can rewrite this reflection identity more explicitly as
\begin{align}
(2\pi)^{-s/2}\zeta(s)\Gamma(s)&=\sqrt{\frac{2}{\pi}}\Gamma(s)\sin(\pi s/2)\cdot (2\pi)^{(s-1)/2}\Gamma(1-s)\zeta(1-s)\\
& =2^{s/2}\pi^{-s/2}\frac{\sin(\pi s/2)}{\pi}\Gamma(s)\Gamma(1-s)\zeta(1-s).
\end{align}We can use the following consequences of the reflection and duplication formula to simplify further:
\[
\frac{\sin(\pi s /2)}{\pi} = \frac{1}{\Gamma(s/2)\Gamma(1 - s/2)}, \quad \frac{\Gamma(1-s)}{\Gamma(1-s/2)} = (2\pi)^{-s}\Gamma((1-s)/2).
\](See appendix B for more detail.) With the help of these identities, we finally obtain
\begin{align}
(2\pi)^{-s/2}\Gamma(s/2)\zeta(s) & = 2^{-s}\pi^{-1/2}\cdot (2\pi)^{s/2}\Gamma((1-s)/2)\zeta(1-s)\\
& = 2^{-s/2}\pi^{(s-1)/2}\Gamma((1-s)/2)\zeta(1-s).
\end{align}Multiplying out $2^{-s/2}$, we see this is equivalent to the functional equation:
\[
\pi^{-s/2}\Gamma(s/2)\zeta(s) = \pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).
\]It is a thing of beauty. But it's very useful too. Not only we can use it to analytically continue $\zeta$, it immediately yields some interesting values. For instance, substitute $s = 2$:
\begin{align*}
\pi^{-1}\Gamma(1)\zeta(2) & = \pi^{1/2}\Gamma(-1/2)\zeta(-1).
\end{align*}Since $\Gamma(-1/2) = -2\Gamma(1/2) = -2\sqrt{\pi}$, and $\zeta(2) = \pi^2/6$, we find that
\[
\zeta(-1) = \frac{\pi^{-3/2}\Gamma(1)\zeta(2)}{\Gamma(-1/2)} = \frac{\sqrt{\pi}}{-12\sqrt{\pi}} = -\frac{1}{12}.
\]
A. Sine transforms
Here, we prove equation $(\ref{sine})$. First, we need the result\[
\frac{1}{e^u - 1} =\frac{1}{u} - \frac{1}{2} + \sum_{k\geq 1}\frac{2u}{4k^2\pi^2 + u^2}.\tag{8}\label{sinlog}
\]To prove this, start with the product formula for sine,
\[
\sin z = z\prod_{k\geq 0}\left(1 -\frac{z^2}{\pi^2k^2}\right).
\]Set $z = u/2i$ and take log of both sides
\[
\log\left[e^{u/2} - e^{-u/2}\right] = \log(1 - e^{-u}) + \frac{u}{2} = \log(u) + \sum_{k\geq 0}\log\left(\frac{4\pi^2k^2 + u^2}{4\pi^2k^2}\right).
\]Differentiating with respect to $u$, we obtain $(\ref{sinlog})$. We can use this to simplify the integral of $(\ref{f(t)})$:
\begin{align*}
\int_0^\infty f(x) \sin(xt)\,dx & = \int_0^\infty \left(\frac{1}{e^{\sqrt{2\pi}x} -1} - \frac{1}{\sqrt{2\pi}x}\right) \sin(xt)\,dx\\
& =\sum_{k\geq 1}\int_0^\infty e^{-kx\sqrt{2\pi}}\sin(xt)\,dx -\frac{1}
{\sqrt{2\pi}}\int_0^\infty \frac{\sin(xt)}{x}\,dx \\
& =\sum_{k\geq 1}\frac{1}{t}\int_0^\infty e^{-(\sqrt{2\pi}k/t)\eta}\sin(\eta)\,d\eta - \frac{1}{4}\sqrt{2\pi} \\
& = \sum_{k\geq 1} \frac{t}{2\pi k^2+t^2}- \frac{1}{4}\sqrt{2\pi}\\
& = \frac{1}{2}\sqrt{2\pi}\left[\frac{1}{e^{\sqrt{2\pi}t} - 1} - \frac{1}{\sqrt{2\pi}t} + \frac{1}{2}\right] -\frac{1}{4}\sqrt{2\pi}\\
& = \sqrt{\frac{\pi}{2}}f(t),
\end{align*}which is equivalent to $(\ref{sine})$. We have used the integrals
\begin{align*}
\int_{0}^\infty \frac{\sin(xt)}{x}\,dx & = \frac{\pi}{2} \\
\int_{0}^\infty e^{-\lambda x}\sin(x)\,dx & = \frac{1}{\lambda^2 + 1}, \quad \lambda > 0.
\end{align*}
B. Gamma function results
1. The reflection formulaTo prove equation $(\ref{gamma})$, namely
\[
\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin \pi s}
\]we recall a convenient way of writing the relation between the gamma functions and beta integral,
\[
\int_0^\infty \frac{t^{\alpha-1}}{(1+t)^{\alpha + \beta}}dt = B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}.
\]For $0 < \alpha < 1$, we set $\beta = 1 - \alpha$ to obtain
\[
\int_0^\infty \frac{t^{\alpha-1}}{1+t}dt = \frac{\Gamma(\alpha)\Gamma(1 - \alpha)}{\Gamma(1)} = \Gamma(\alpha)\Gamma(1 - \alpha).
\]It is now just a matter of doing the integral. Consider the contour integral
\[
\int_{C(R, \epsilon)} \frac{z^{\alpha - 1}}{1 - z} dz,
\]where $C(R, \epsilon)$ is the following contour:
The only singularity within the contour lies at $z = 1$. Provided we take the principal value of $z^{\alpha - 1}$, the residue is simply $-1^{\alpha - 1} = -1$. By the residue theorem,
\[
\int_{C(R, \epsilon)} \frac{z^{\alpha - 1}}{1 - z} dz = -2\pi i.
\]We split the integral up as in the picture, and note that along $C_1$, $z = Re^{i\theta}$; along $C_2$, $z = e^{i\pi}t$; along $C_3$, $z = z = \epsilon e^{i\theta}$; and along $C_4$, $z = e^{-i\pi}t$. (The $e^{\pm i\pi}$ is significant, and dictates how we take powers consistent with the principal value used to compute the residue.) Hence,
\begin{align}
-2\pi i & = \int_{C_1} \frac{z^{\alpha - 1}}{1 - z} dz + \int_{C_2} \frac{z^{\alpha - 1}}{1 - z} dz + \int_{C_3} \frac{z^{\alpha - 1}}{1 - z} dz + \int_{C_4} \frac{z^{\alpha - 1}}{1 - z} \\
& = \int_{-\pi}^\pi \frac{iR^{\alpha}e^{i\alpha\theta}}{1 - Re^{i\theta}} d\theta + \int_R^\epsilon \frac{t^{\alpha - 1}e^{i\alpha\pi}}{1 + t} dt + \int_{\pi}^{-\pi} \frac{i\epsilon^{\alpha}e^{i\alpha\theta}}{1 - \epsilon e^{i\theta}} d\theta + \int_\epsilon^R \frac{t^{\alpha - 1}e^{-i\alpha\pi}}{1 + t} dt \\
& = (e^{-i\alpha\pi} - e^{i\alpha\pi})\int_\epsilon^R \frac{t^{\alpha - 1}}{1 + t} dt +\int_{-\pi}^\pi \frac{iR^{\alpha}e^{i\alpha\theta}}{1 - Re^{i\theta}} d\theta + \int_{\pi}^{-\pi} \frac{i\epsilon^{\alpha}e^{i\alpha\theta}}{1 - \epsilon e^{i\theta}} d\theta.
\end{align}If we send $\epsilon \to 0$ and $R \to \infty$, the $C_1$ and $C_3$ angular integrals vanish, since the $C_1$ integrand $\sim R^\alpha / R = R^{\alpha-1}$ and the $C_3$ integrand $\sim \epsilon^\alpha$. Thus
\begin{align*}
2\pi i & = (e^{i\alpha\pi} - e^{-i\alpha\pi})\int_\epsilon^R \frac{t^{\alpha - 1}}{1 + t} dt \\
& = 2 i \sin(\alpha \pi)\int_\epsilon^R \frac{t^{\alpha - 1}}{1 + t} dt = 2 i \sin(\alpha \pi)\Gamma(\alpha)\Gamma(1 - \alpha).
\end{align*}The reflection formula $(\ref{gamma})$ follows for all complex $\alpha$ by analytic continuation. (For the singular points $\alpha \in \mathbb{Z}_{\leq 0}$, the Laurent series coefficients also match.)
2. The duplication formula
We now prove the duplication formula $(\ref{gamma2})$. First, we make the substitution $t = (1+u)/2$ in a more traditional expression for the beta function:
\begin{align}
B(z, z) = \frac{\Gamma(z)\Gamma(z)}{\Gamma(2z)} & = \int_0^1 t^{z - 1}(1-t)^{z-1} \,dt \\
&= \frac{1}{2}\int_{-1}^1 \left(\frac{1+u}{2}\right)^{z-1}\left(\frac{1-u}{2}\right)^{z-1} \,du\\
& =\frac{1}{2^{2z-1}}\int_{-1}^1 (1-u^2)^{z-1} \,du = \frac{1}{2^{2z-2}}\int_{0}^1 (1-u^2)^{z-1} \,du.
\end{align}But if we make the substitution $t = u^2$ in $B(1/2, z)$, we get
\begin{align}
B(\tfrac{1}{2}, z) = \frac{\Gamma(\tfrac{1}{2})\Gamma(z)}{\Gamma(z + \tfrac{1}{2})} & = \int_0^1 t^{1/2 - 1}(1-t)^{z-1}\,dt \\
& = \int_0^1 u^{- 1}(1-u^2)^{z-1}2u\,du \\
& = 2\int_0^1 (1-u^2)^{z-1}\,du.
\end{align}Equating the integrals yields
\[
2^{2z-1}\frac{\Gamma(z)\Gamma(z)}{\Gamma(2z)} = \frac{\Gamma(\tfrac{1}{2})\Gamma(z)}{\Gamma(z + \tfrac{1}{2})}.
\]Using $\Gamma(1/2) = \sqrt{\pi}$ and rearranging, we finally obtain $(\ref{gamma2})$:
\[
\Gamma(z)\Gamma(z+\tfrac{1}{2}) = 2^{1-2z}\sqrt{\pi}\Gamma(2z).
\] An integral
Finally, we sketch a proof that for $x > 0$,
\begin{align}
\int_0^\infty y^{s-1} \sin(xy)\,dy & = x^{-s}\Gamma(s)\sin(\pi s/2).
\end{align} Suppose that $0 < s < 1$. Making the change of variable $t = xy$,
\begin{align}
\int_0^\infty y^{s-1} \sin(xy)\,dy & = x^{-s}\int_0^\infty t^{s-1} \sin(t)\,dt\\
& = x^{-s}\Im \int_0^\infty t^{s-1} e^{it}\,dt.
\end{align}This looks like the gamma integral, except for the factor of $-i$ in the exponential. If we integrate around the boundary $C_R$ of the first quadrant of the disc of size $R$, we find from the residue theorem that
\begin{align*}
\int_0^R z^{s-1} e^{iz}\,dz + I(R) + \int_{iR}^0 z^{s-1} e^{iz}\,dz = 0,
\end{align*}where
\[
I(R) \sim R^{s-1}\int e^{-R\sin\theta} d\theta \to 0
\]as $R\to\infty$. Hence,
\begin{align*}
\int_0^\infty z^{s-1} e^{iz}\,dz &= \lim_{R\to\infty}\int_0^{iR} z^{s-1} e^{iz}\,dz\\
&= i^s\lim_{R\to\infty}\int_{0}^{R} x^{s-1} e^{-x}\,dx \\
&= e^{i\pi s/2}\Gamma(s).
\end{align*}Taking the imaginary component gives the desired result. By analytic continuation, the result is true for $s = \sigma + i\tau$, $0 < \sigma < 1$.
References
- The Riemann Zeta-Function: Theory and Applications (1985), Aleksander Ivić.
- Special functions and their symmetries, Vadim Kuznetsov.
- Legendre's duplication formula, Proof Wiki.
Written on May 17, 2015